(i) The three conditions for a binomial distribution are:

1. The probability of success, denoted as \(p\), is constant for each trial.
2. The trials are independent.
3. There are a fixed number of trials, and each trial has only two possible outcomes (success or failure).

(ii) Let \(X\) be the number of months George buys shares in a small company. \(X\) follows a binomial distribution with parameters \(n = 18\) and \(p = 0.15\).

We need to find \(P(X \geq 3)\), which is equivalent to \(1 - P(X \leq 2)\).

Calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\) using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

\(P(X = 0) = \binom{18}{0} (0.15)^0 (0.85)^{18} = (0.85)^{18}\)

\(P(X = 1) = \binom{18}{1} (0.15)^1 (0.85)^{17} = 18 \times (0.15) \times (0.85)^{17}\)

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16} = \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}\)

Thus, \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(P(X \geq 3) = 1 - P(X \leq 2)\)

Substitute the values:

\(P(X \geq 3) = 1 - [(0.85)^{18} + 18 \times (0.15) \times (0.85)^{17} + \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}]\)

\(P(X \geq 3) = 0.520\)